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Q: 11.12 (b) Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of 56\hspace{1mm}V.

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De Broglie wavelength is given by the De Broglie relation as

\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{4.038\times 10^{-24}}\\ \lambda =0.164\ nm

the wavelength is 0.164 nm

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